Question: Evaluate the double integral. $ \int_{-\pi}^0 \int_{\sin(x)}^{2} 3 + \cos(x) \, dy \, dx =$ Choose 1 answer: Choose 1 answer: (Choice A) A $-8\pi + 4$ (Choice B) B $4\pi + 6$ (Choice C) C $6\pi + 6$ (Choice D) D $2\pi + 4$
Explanation: First, we evaluate the inner integral. We can substitute in the $\sin(x)$ at the end as if it were a numerical bound. $\begin{aligned} & \int_{-\pi}^0 \int_{\sin(x)}^{2} 3 + \cos(x) \, dy \, dx \\ \\ &= \int_{-\pi}^0 \left[ 3y + y\cos(x) \right]_{\sin(x)}^2 dx \\ \\ &= \int_{-\pi}^0 6 + 2\cos(x) - 3\sin(x) - \sin(x)\cos(x) \, dx \end{aligned}$ Second, we evaluate the outer integral. We can use the trigonometric identity $\sin(2x) = 2\sin(x)\cos(x)$ to integrate $\sin(x)\cos(x)$. $\begin{aligned} &\int_{-\pi}^0 6 + 2\cos(x) - 3\sin(x) - \sin(x)\cos(x) \, dx \\ \\ &= \left[ 6x + 2\sin(x) + 3\cos(x) + \dfrac{1}{4} \cos(2x) \right]_{-\pi}^0 \\ \\ &= 3 (1) + \dfrac{1}{4} - \left( -6\pi + 3(-1) + \dfrac{1}{4} \right) \\ \\ &= 6\pi + 6 \end{aligned}$ The answer: $ \int_{-\pi}^0 \int_{\sin(x)}^{2} 3 + \cos(x) \, dy \, dx = 6\pi + 6$